Norms are equivalent iff they induce the same topology

Let $(𝑋, 𝕂)$ be a vector space and $𝑝, 𝑞 : 𝑋 \to ℝ$ be norms. Let $T 𝑝$ and $T 𝑞$ be the topologies on $𝑋$ induced by $𝑝$ and $𝑞$ respectively. Then $𝑝$ and $𝑞$ are equivalent norms iff $T 𝑝 = T 𝑞$ . topology

Proof

First suppose that $𝑝$ and $𝑞$ are equivalent, i.e. there exist $𝑏 \geq 𝑎 > 0$ such that $𝑎 𝑞 (𝑥) \leq 𝑝 (𝑥) \leq 𝑏 𝑞 (𝑥)$ for all $𝑥 \in 𝑋$ . Now suppose $𝑈 \subseteq 𝑋$ is open under $𝑝$ . Then for every $𝑥 \in 𝑈$ there exists some $𝛿 > 0$ such that $𝑝 (𝑦 - 𝑥) < 𝛿 ⟹ 𝑦 \in 𝑈$ . But $𝑝 (𝑦 - 𝑥) \leq 𝑏 𝑞 (𝑦 - 𝑥)$ , and thus for every $𝑥 \in 𝑈$ there exists $𝛿 𝑏 > 0$ such that $𝑞 (𝑦 - 𝑥) < 𝛿 𝑏 ⟹ 𝑝 (𝑦 - 𝑥) < 𝛿 ⟹ 𝑦 \in 𝑈$ . Hence $𝑈$ is open under $𝑞$ . Therefore $T 𝑝 \subseteq T 𝑞$ Since equivalence of norms is symmetric, by the same argument $T 𝑞 \subseteq T 𝑝$ , and thus $T 𝑝 = T 𝑞$ .

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Norms are equivalent iff they induce the same topology

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